On the Derived Length of Lie Solvable Group Algebras

20 CHAPTER 3
this bound is always achieved for odd p, that is, Proposition 3.1.1 is
valid for arbitrary nilpotent groups with cyclic commutator subgroup,
provided that p is odd.
Theorem 3.1.2. Let G be a nilpotent group with cyclic commutator
subgroup of order p n , where p is an odd prime, and let char(F ) = p.
Then
dlL(F G) = dl L (F G) = ⌈log 2(p n + 1)⌉.
Proof. Let G ′ = 〈x | xpn = 1〉 and let us choose a, b ∈ G such that
x = (a, b). First of all, we claim that
(3.1) [b l a m , b s a t ] ≡ (ms − lt)b l+s a m+t (x − 1) (mod I(G ′ ) 2 )
for every l, s, m, t ∈ Z. Indeed, an easy computation yields
b l a m , b s a t = b s a t b l a m (b l a m , b s a t ) − 1
(3.2)
and since now γ3(G) ⊆ (G ′ ) p ,
= b l+s a m+t (a t , b l ) am (b l a m , b s a t ) − 1
≡ b l+s a m+t (b l a m , b s a t ) − 1
(mod I(G ′ ) 2 ),
(b l a m , b s a t ) ≡ (b l , a t )(a m , b s )
≡ (b, a) lt (a, b) ms ≡ x ms−lt
(mod (G ′ ) p ).
Thus (b l a m , b s a t ) = x ms−lt+pi for some i. In view of the identity
we have
uv − 1 = (u − 1)(v − 1) + (u − 1) + (v − 1),
(b l a m , b s a t ) − 1 ≡ (ms − lt + pi)(x − 1)
≡ (ms − lt)(x − 1) (mod I(G ′ ) 2 )
and putting this into (3.2) we obtain (3.1).
Now, let k ≥ 1, l, m, s, t ∈ Z, z1, z2 ∈ I(G ′ ) 2k
and set
fk(l, m, s, t, z1, z2) = b l a m (x − 1) 2k−1 + z1, b s a t (x − 1) 2k −1
+ z2 .